The Unapologetic Mathematician

Mathematics for the interested outsider

HEY YOU GUYS!

They’re gonna turn it back on!

May 14, 2008 Posted by John Armstrong | Uncategorized | | 3 Comments

Oh, Flock

I’m testing a new browser: Flock Partly this is because it’s supposed to have a built-in WordPress editor, which I’m trying to use now.  From the looks of it, though, it really doesn’t have much beyond bare-bones support, which just makes it good for little notes like this, but not proper posts like I make.

I’ll give the browser a try, but I think I’ll end up falling back to Safari.

May 13, 2008 Posted by John Armstrong | Uncategorized | | 8 Comments

60 Second Science?

A post just went up on the Scientific American website about the “McGurck effect”. Supposedly when you hear someone say the syllable “ba” and see them say “ga”, you’ll think you’re hearing the syllable “da”. They’ve even got a video to try it for yourself.

And even though you now know it’s an illusion—you will still, when you see the video, think you are hearing “da”.

Except it doesn’t work. So I called my mother over and didn’t tell her what was supposed to happen. Just showed the video and asked what syllable she heard. Right away, with no hesitation, she said he was saying “ba” but the lips were moving like “ga”. Not even a trace of the desired effect.

So are we mutants? Try it yourself. Try it on your unsuspecting friends and family members and ask them what they hear. And tell me what happens. Remember: it’s not science unless we can falsify it.

[UPDATE]: As I mention in the comments, I found the source of that video. There they say that the effect shows up in 98% of adults. So my mother and I are evidently among the 1/50 of adult humans who can separate visual and auditory inputs inside our heads.

May 13, 2008 Posted by John Armstrong | Uncategorized | | 8 Comments

Commutativity in Series III

Okay, here’s the part I promised I’d finish last Friday. How do we deal with rearrangements that “go to infinity” more than once? That is, we chop up the infinite set of natural numbers into a bunch of other infinite sets, add each of these subseries up, and then add the results up. If the original series was absolutely convergent, we’ll get the same answer.

First of all, if a series \sum_{k=0}^\infty a_k converges absolutely, then so does any subseries \sum_{j=0}^\infty a_{p(j)}, where p is an injective (but not necessarily bijective!) function from the natural numbers to themselves. For instance, we could let p(j)=2j and add up all the even terms from the original series.

To see this, notice that at any finite n we have a maximum value N=\max\limits_{0\leq j\leq n}p(j). Then we find

\displaystyle\left|\sum\limits_{j=0}^na_{p(j)}\right|\leq\sum\limits_{j=0}^n\left|a_{p(j)}\right|\leq\sum\limits_{k=0}^N\left|a_k\right|\leq\sum\limits_{k=0}^\infty\left|a_k\right|

So the new sequence of partial sums of absolute values is increasing and bounded above, and thus converges.

Now let’s let p_0, p_1, p_2, and so on be a countable collection of functions defined on the natural numbers. We ask that

  • Each p_n is injective.
  • The image of p_n is a subset P_k\subseteq\mathbb{N}.
  • The collection \left\{P_0,P_1,P_2,...\right\} is a partition of \mathbb{N}. That is, these subsets are mutually disjoint, and their union is all of \mathbb{N}.

If \sum_{k=0}^\infty a_k is an absolutely convergent series, we define \left(b_n\right)_j=a_{p_n(j)} — the subseries defined by p_n. Then from what we said above, each \sum_{j=0}^\infty\left(b_n\right)_j is an absolutely convergent series whose sum we call s_n. We assert now that \sum_{n=0}^\infty s_n is an absolutely convergent series whose sum is the same as that of \sum_{k=0}^\infty a_k.

Let’s set t_m=\sum_{n=0}^m\left|s_n\right|. That is, we have

\displaystyle t_m\leq\sum\limits_{j=0}^\infty\left|\left(b_1\right)_j\right|+...+\sum\limits_{j=0}^\infty\left|\left(b_m\right)_j\right|=\sum\limits_{j=0}^\infty\left(\left|\left(b_1\right)_j\right|+...+\left|\left(b_m\right)_j\right|\right)

But this is just the sum of a bunch of absolute values from the original series, and so is bounded by \sum_{k=0}^\infty\left|a_k\right|. So the series of absolute values of s_n has bounded partial sums, and so \sum_{n=0}^\infty s_n converges absolutely. That it has the same sum as the original is another argument exactly analogous to (but more complicated than) the one for a simple rearrangement, and for associativity of absolutely convergent series.

This pretty much wraps up all I want to say about calculus for now. I’m going to take a little time to regroup before I dive into linear algebra in more detail than the abstract algebra I covered before. But if you want to get ahead, go back and look over what I said about rings and modules. A lot of that will be revisited and fleshed out in the next sections.

May 12, 2008 Posted by John Armstrong | Analysis, Calculus | | No Comments

Sunday Samples 68

Like last year, we come again to a difficult week. I think I’ll call back to the late ’90s: days of the Lilith Fair (yes, I went to all three). Track one, side one of Tracy Bonham’s major-label debut The Burdens of Being Upright is the appropriately-titled “Mother Mother”.
Read more »

May 11, 2008 Posted by John Armstrong | Sunday Samples | | No Comments

Commutativity in Series II

We’ve seen that commutativity fails for conditionally convergent series. It turns out, though, that things are much nicer for absolutely convergent series. Any rearrangement of an absolutely convergent series is again absolutely convergent, and to the same limit.

Let \sum_{k=0}^\infty a_k be an absolutely convergent series, and let p:\mathbb{N}\rightarrow\mathbb{N} be a bijection. Define the rearrangement b_k=a_{p(k)}.

Now given an \epsilon>0, absolute convergence tells us we can pick an N so that any tail of the series of absolute values past that point is small. That is, for any n\geq N we have

\displaystyle\sum\limits_{k=n+1}^\infty\left|a_k\right|<\frac{\epsilon}{2}

Now for 0\leq n\leq N, the function p^{-1} takes only a finite number of values (the inverse function exists because p is a bijection). Let M be the largest such value. Thus if m>M we will know that p(m)>N. Then for any such m we have

\displaystyle\sum\limits_{j=m+1}^{m+d}\left|b_k\right|=\sum\limits_{j=m+1}^{m+d}\left|a_{p(k)}\right|\leq\sum\limits_{k=N+1}^\infty\left|a_k\right|

and we know that the sum on the right is finite by the assumption of absolute convergence. Thus the tail of the series of b_j — and thus the series itself — must converge. Now a similar argument to the one we used when we talked about associativity for absolutely convergent series shows that the rearranged series has the same sum as the original.

This is well and good, but it still misses something. We can’t handle reorderings that break up the order structure. For example, we might ask to add up all the odd terms, and then all the even terms. There is no bijection p that handles this situation. And yet we can still make it work.

Unfortunately, I arrive in Maryland having left my references back in New Orleans. For now, I’ll simply assert that for absolutely convergent series we can perform these more general rearrangements, though I’ll patch this sometime.

May 9, 2008 Posted by John Armstrong | Analysis, Calculus | | 1 Comment

Commutativity in Series I

We’ve seen that associativity may or may not hold for infinite sums, but it can be improved with extra assumptions. As it happens, commutativity breaks down as well, though the story is a bit clearer here.

First we should be clear about what we’re doing. When we add up a finite list of real numbers, we can reorder the list in many ways. In fact, reorderings of n numbers form the symmetric group S_n. If we look back at our group theory, we see that we can write any element in this group as a product of transpositions which swap neighboring entries in the list. Thus since the sum of two numbers is invariant under such a swap — a+b=b+a — we can then rearrange any finite list of numbers and get the same sum every time.

Now we’re not concerned about finite sums, but about infinite sums. As such, we consider all possible rearrangements — bijections p:\mathbb{N}\rightarrow\mathbb{N} — which make up the “infinity symmetric group S_\infty. Now we might not be able to effect every rearrangement by a finite number of transpositions, and commutativity might break down.

If we have a series with terms a_k and a bijection p, then we say that the series with terms b_k=a_{p(k)} is a rearrangement of the first series. If, on the other hand, p is merely injective, then we say that the new series is a subseries of the first one.

Now, if \sum_{k=0}^\infty a_k is only conditionally convergent, I say that we can rearrange the series to give any value we want! In fact, given x\leq y (where these could also be \pm\infty) there will be a rearrangement b_k=a_{p(k)} so that

\displaystyle\liminf\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=x
\displaystyle\limsup\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=y

First we throw away any zero terms in the series, since those won’t affect questions of convergence, or the value of the series if it does converge. Then let p_n be the nth positive term in the sequence a_k, and let -q_n be the nth negative term.

The two series with positive terms \sum_{k=0}^\infty p_k and \sum_{k=0}^\infty q_k both diverge. Indeed, if one converged but the other did not, then the original series \sum_{k=0}^\infty a_k would diverge. On the other hand, if they both converged then the original series would converge absolutely. Conditional convergence happens when the subseries of positive terms and the subseries of negative terms just manage to balance each other out.

Now we take two sequences x_n and y_n converging to x and y respectively. Since the series of positive terms diverges, they’ll eventually exceed any positive number. We can take just enough of them (say k_1 so that

\displaystyle\sum_{k=0}^{k_1}p_k>y_1

Similarly, we can then take just enough negative terms so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l<x_1

Now take just enough of the remaining positive terms so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k>y_2

and enough negatives so that

\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k-\sum\limits_{l=l_1+1}^{l_2}q_l<x_2

and so on and so forth. This gives us a rearrangement of the terms of the series.

Each time we add positive terms we come within p_{k_j} of y_j, and each time we add negative terms we come within q_{l_j} of x_j. But since the original sequence a_n must be converging to zero (otherwise the series couldn’t converge), so must the p_{k_j} and q_{l_j} be converging to zero. And the sequences x_j and y_j are converging to x and y.

It’s straightforward from here to show that the limits superior and inferior of the partial sums of the rearranged series are as we claim. In particular, we can set them both equal to the same number and get that number as the sum of the rearranged series. So for conditionally convergent series, the commutativity property falls apart most drastically.

May 8, 2008 Posted by John Armstrong | Analysis, Calculus | | 2 Comments

Quantum Knot Mosaics

Today, Sam Lomonaco and Louis Kauffman posted to the arXiv a paper on “Quantum Knots and Mosaics”. I had the pleasure of a sneak preview back in March. Here’s what I said then (I haven’t had a chance to read the paper as posted, so some of this may be addressed):

About half the paper consists of setting up definitions of a mosaic and the Reidemeister moves. This concludes with the conjecture that before you allow superpositions the mosaic framework captures all of knot theory.

The grading by the size of the mosaic leads to an obvious conjecture: there exist mosaic knots which are mosaic equivalent, but which require arbitrarily many expansions. This is analogous to the same fact about crossing numbers.

Obviously, I’d write these combinatorial frameworks as categories with the mosaics as objects and the morphisms generated by the mosaic moves. Superpositions just seem to be the usual passage from a set to the vector space on that basis. See my new paper for how I say this for regular knots and Reidemeister moves.

Then (like I say in the paper) we want to talk about mosaic “covariants”. I think this ends up giving your notion of invariant after we decategorify (identify isomorphic outputs).

The only thing I’m wondering about (stopping shy of saying you two are “wrong”) is the quantum moves. The natural thing would be to go from the “group” (really its a groupoid like I said before) of moves to its linearization. That is, we should allow the “sum” of two moves as a move. This splits a basis mosaic input into a superposition.

In particular, the “surprising” result you state that one quantum mosaic is not quantum equivalent to the other must be altered. There is clearly a move in my view taking the left to the right. “Equivalence” is then the statement that two quantum mosaics are connected by an *invertible* move. I’m not sure that the move from left to right is invertible yet, but I think it is.

May 8, 2008 Posted by John Armstrong | Knot theory | | No Comments

Associativity in Series II

I’m leaving for DC soon, and may not have internet access all day. So you get this now!

We’ve seen that associativity doesn’t hold for infinite sums the way it does for finite sums. We can always “add parentheses” to a convergent sequence, but we can’t always remove them.

The first example we mentioned last time. Consider the series with terms a_k=(-1)^k:

\displaystyle\sum\limits_{k=0}^\infty a_k=1+(-1)+1+(-1)+...

Now let’s add parentheses using the sequence d(j)=2j+1. Then b_j=(-1)^{(2(j-1)+1)+1}+(-1)^{2j+1}=1+(-1)=0. That is, we now have the sequence

\displaystyle\sum\limits_{j=0}^\infty b_j=(1+(-1))+(1+(-1))+...=0+0+...=0

So the resulting series does converge. However, the original series can’t converge.

The obvious fault is that the terms a_k don’t get smaller. And we know that \lim\limits_{k\rightarrow\infty}a_k must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms b_j go to zero, but when we remove them this condition can fail. And it turns out there’s just one more condition we need so that we can remove parentheses.

So let’s consider the two series with terms a_k and b_j, where the first is obtained from the second by removing parentheses using the function d(j). Assume that \lim_{k\rightarrow\infty}a_k=0, and also that there is some M>0 so that each of the b_j is a sum of fewer than M of the a_k. That is, d(j+1)-d(j)<M. Then the series either both diverge or both converge, and if they converge they have the same sum.

We set up the sequences of partial sums

\displaystyle s_n=\sum\limits_{k=0}^na_k
\displaystyle t_m=\sum\limits_{j=0}^mb_j

We know from last time that t_m=s_{d(m)}, and so if the first series converges then the second one must as well. We need to show that if t=\lim\limits_{m\rightarrow\infty}t_m exists, then we also have \lim\limits_{n\rightarrow\infty}s_n=t.

To this end, pick an \epsilon>0. Since the sequence of t_m converge to t, we can choose some N so that \left|t_m-t\right|<\frac{\epsilon}{2} for all m>N. Since the sequence of terms a_k converges to zero, we can increase N until we also have \left|a_k\right|<\frac{\epsilon}{2M} for all k>N.

Now take any n>d(N). Then n falls between d(m) and d(m+1) for some m. We can see that m\geq N, and that n is definitely above N. So the partial sum s_n is the sum of all the a_k up through k=d(m+1), minus those terms past k=n. That is

\displaystyle s_n=\sum\limits_{k=0}^na_k=\sum\limits_{k=0}^{d(m+1)}a_k-\sum\limits_{n+1}^{d(m+1)}a_k

But this first sum is just the partial sum t_{m+1}, while each term of the second sum is bounded in size by our assumptions above. We check

\displaystyle\left|s_n-t\right|=\left|(t_{m+1}-t)-\sum\limits_{n+1}^{d(m+1)}a_k\right|\leq\left|t_{m+1}-t\right|+\sum\limits_{n+1}^{d(m+1)}\left|a_k\right|

But since n is between d(m) and d(m+1), there must be fewer than M terms in this last sum, all of which are bounded by \frac{\epsilon}{2M}. So we see

\displaystyle\left|s_n-t\right|<\frac{\epsilon}{2}+M\frac{\epsilon}{2M}=\epsilon

and thus we have established the limit.

May 7, 2008 Posted by John Armstrong | Analysis, Calculus | | 3 Comments

Associativity in Series I

As we’ve said before, the real numbers are a topological field. The fact that it’s a field means, among other things, that it comes equipped with an associative notion of addition. That is, for any finite sum we can change the order in which we perform the additions (though not the order of the terms themselves — that’s commutativity).

The topology of the real numbers means we can set up sums of longer and longer sequences of terms and talk sensibly about whether these sums — these series — converge or not. Unfortunately, this topological concept ends up breaking the algebraic structure in some cases. We no longer have the same freedom to change the order of summations.

When we write down a series, we’re implicitly including parentheses all the way to the left. Consider the partial sums:

\displaystyle s_n=\sum\limits_{k=0}^na_k=((...(((a_0+a_1)+a_2)+a_3)...+a_{n-1})+a_n)

But what if we wanted to add up the terms in a different order? Say we want to write

\displaystyle s_6=(((a_0+a_1)+(a_2+a_3))+((a_4+a_5)+a_6))

Well this is still a left-parenthesized expression, it’s just that the terms are not the ones we looked at before. If we write b_0=a_0+a_1, b_1=a_2+a_3, and b_2=a_4+a_5+a_6 then we have

\displaystyle s_6=((b_0+b_1)+b_2)=\sum\limits_{j=0}^2b_j=t_2

So this is actually a partial sum of a different (though related) series whose terms are finite sums of terms from the first series.

More specifically, let’s choose a sequence of stopping points: an increasing sequence of natural numbers d(j). In the example above we have d(0)=1, d(1)=3, and d(3)=6. Now we can define a new sequence

\displaystyle b_0=\sum\limits_{k=0}^{d(0)}a_k
\displaystyle b_j=\sum\limits_{k=d(j-1)+1}^{d(j)}a_k

Then the sequence of partial sums t_m of this series is a subsequence of the s_n. Specifically

\displaystyle t_m=\sum\limits_{j=0}^mb_j=\sum\limits_{k=0}^{d(0)}a_k+\sum\limits_{j=1}^m\left(\sum\limits_{k=d(j-1)+1}^{d(j)}a_k\right)=\sum\limits_{k=0}^{d(m)}a_k=s_{d(m)}

We say that the sequence t_m is obtained from the sequence s_n by “adding parentheses” (most clearly notable in the above expression for t_m). Alternately, we say that s_n is obtained from t_m by “removing parentheses”.

If the sequence s_n converges, so must the subsequence t_m=s_{d(m)}, and moreover to the same limit. That is, if the series \sum_{k=0}^\infty a_k converges to s, then any series \sum_{j=0}^\infty b_j obtained by adding parentheses also converges to s.

However, convergence of a subsequence doesn’t imply convergence of the sequence. For example, consider a_k=(-1)^k and use d(j)=2j+1. Then s_n jumps back and forth between zero and one, but t_m is identically zero. So just because a series converges, another one obtained by removing parentheses may not converge.

May 6, 2008 Posted by John Armstrong | Analysis, Calculus | | 2 Comments

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